In the previous section we saw the loads on the flight of a stair which is supported on landings. In this section we will see the loads on the landings.
Part 2: Loads on the Landings:
In part 1, we have calculated w2. We have seen that it is calculated by the same method that we would use for calculating the w2 of an ordinary longitudinal stair. (See Eq.16.10). This is the original load on the landing. It is obtained on a 1 x 1 m square area.
There is more. w2 calculated above is the original load on the landing. But the landing is supporting the flight also. So the load from the 'Going' will also come on to the landing. Let us see how this load from the Going is distributed on the landing.
We have the load w1 on the Going. It is the load per 1 x 1 m sq. area (horizontal projection). So, once we calculate w1, we will be able to calculate the total load on the going. From the fig.16.41 below, this will be equal to w1 x Fw x l1.
Fig.16.41
Transfer of loads from the Going to the Landing
Where Fw is the width of the flight and l1 is the clear length of the Going. This is the total load acting in the total length of the Going. Half of this will be transferred to one landing and the other half to the other landing. So in the fig.16.41, the load from pqrs will be transferred to the landing 1. Similarly the load from qruy will be transferred to the landing 2.
So now we know the magnitude of the load which is transferred from the Going to the landing. Next we must know how this much load is distributed on the landing. This is also shown in the fig 16.41. The load from pqrs is assumed to be distributed uniformly on the corresponding area pmns of the landing. As the load is uniformly distributed, we can divide the load by the area to get the load per 1 x 1 m sq.
• Load on pqrs = 0.5w1Fw l1
• Area of mnsp = Fw Lw .(Where Lw is the width of the landing).
So load per unit area =
This is the load on one half (mnsp) of the landing. The other half will also be subjected to the same load because it is supporting the other flight. Thus the whole landing is carrying the load 0.5w1l1 / Lw.
So we have obtained the two loads:
• The original load on the landing = [w2] per sq.mt
• The load coming from the Going = [0.5w1l1 / Lw] per sq.mt
The sum of the two will be the total load acting on the landing per sq.mt. Thus:
Eq.16.19: The total load acting on the landing per sq.mt =
What role does this 'total load' play in the analysis and design of the landing slab? Let us find out:
The landing slab is designed as a One-way slab. It spans between the hatched portions shown in fig.16.41. We have already seen the design of one-way slabs here. We have seen that the load on a 1 x 1m square area of the slab is the same UDL per meter length, that is required for the design of a 1 m wide strip. So we conclude that, the above 'total load' is the UDL per meter length that we require to design a 1m wide strip of our landing slab.
So we have the methods to obtain all the loads on the stair. With these loads, we can do the analysis to find the Bending moments and shear forces. Once they are calculated, we can do the design.
We will now do the analysis and design of our stair. It is given as:
Solved example 16.2
The arrangement of bars according to the design in the above solved example is shown in the figs.16.42 given below:
Fig.16.42
Reinforcement details of Going and Landing
Now we will discuss some details about the above fig.
As in previous examples, bar ‘a’ becomes top bar in the top landing, and to compensate for this, bar ‘b’ is provided at the same diameter and spacing as ‘a’. Also, ‘a’ and ‘b’ are given enough embedment as explained earlier on the basis of fig.16.25. At the bottom intermediate landing, ‘c’ and ‘d’ are provided to resist any possible hogging moment. When there is hogging moment, these bars will be in tension, and so in order to prevent them from straightening up, they are given enough embedment in accordance with fig.16.25.
One more type of bar, named as ‘e’ is provided at the top landing. The reason for providing these bars can be explained as follows: The stair is supported on the landings. The support for the flight comes from the interior portion of the Landing. But the inner edge of the Landing (the line of intersection between horizontal landing portion and inclined flight portion) is also capable of providing some support. So the edge will act as an intermediate support of a continuous system. We know that hogging moments will develop at intermediate supports, and so top steel will have to be provided. Thus ‘e’ is provided to resist this hogging moment.
Another point that we have to note is the position of the main bars of the Landings. We can see that it is provided as the bottom most layer. This will give maximum effective depth, and so the ‘Moment of resistance’ MuR of the section of the Landing will be more. But in any case , these bars should indeed be the bottom most layer because, they have to provide support to the stair by carrying the bars(‘a’, ‘b’, ‘c’, ‘d’ and ‘e’) of the stairs. And in order to carry them, the bars of the Landing should be the bottom most layer.
Distributor bars for the Landing need not be designed because the bars of the stairs are coming in a transverse direction to the main bars of the Landing.
A small amount of steel (#8 @ 200 c/c) is also provided as top bars for the Landings. These bars will resist any hogging moment developed at the supports (the hatched portions in fig.16.41 above) of the Landings, and will also assist in tying the top bars of stairs, and thus to keep them in position.
It may also be noted that in the above fig., bars of different types are shown separately only for clarity. This was explained earlier based on the fig.16.25(b).
In the next section, we will discuss the load distribution in a particular type of longitudinal stairs.
Part 2: Loads on the Landings:
In part 1, we have calculated w2. We have seen that it is calculated by the same method that we would use for calculating the w2 of an ordinary longitudinal stair. (See Eq.16.10). This is the original load on the landing. It is obtained on a 1 x 1 m square area.
There is more. w2 calculated above is the original load on the landing. But the landing is supporting the flight also. So the load from the 'Going' will also come on to the landing. Let us see how this load from the Going is distributed on the landing.
We have the load w1 on the Going. It is the load per 1 x 1 m sq. area (horizontal projection). So, once we calculate w1, we will be able to calculate the total load on the going. From the fig.16.41 below, this will be equal to w1 x Fw x l1.
Fig.16.41
Transfer of loads from the Going to the Landing
Where Fw is the width of the flight and l1 is the clear length of the Going. This is the total load acting in the total length of the Going. Half of this will be transferred to one landing and the other half to the other landing. So in the fig.16.41, the load from pqrs will be transferred to the landing 1. Similarly the load from qruy will be transferred to the landing 2.
So now we know the magnitude of the load which is transferred from the Going to the landing. Next we must know how this much load is distributed on the landing. This is also shown in the fig 16.41. The load from pqrs is assumed to be distributed uniformly on the corresponding area pmns of the landing. As the load is uniformly distributed, we can divide the load by the area to get the load per 1 x 1 m sq.
• Load on pqrs = 0.5w1Fw l1
• Area of mnsp = Fw Lw .(Where Lw is the width of the landing).
So load per unit area =
This is the load on one half (mnsp) of the landing. The other half will also be subjected to the same load because it is supporting the other flight. Thus the whole landing is carrying the load 0.5w1l1 / Lw.
So we have obtained the two loads:
• The original load on the landing = [w2] per sq.mt
• The load coming from the Going = [0.5w1l1 / Lw] per sq.mt
The sum of the two will be the total load acting on the landing per sq.mt. Thus:
Eq.16.19: The total load acting on the landing per sq.mt =
The landing slab is designed as a One-way slab. It spans between the hatched portions shown in fig.16.41. We have already seen the design of one-way slabs here. We have seen that the load on a 1 x 1m square area of the slab is the same UDL per meter length, that is required for the design of a 1 m wide strip. So we conclude that, the above 'total load' is the UDL per meter length that we require to design a 1m wide strip of our landing slab.
So we have the methods to obtain all the loads on the stair. With these loads, we can do the analysis to find the Bending moments and shear forces. Once they are calculated, we can do the design.
We will now do the analysis and design of our stair. It is given as:
Solved example 16.2
The arrangement of bars according to the design in the above solved example is shown in the figs.16.42 given below:
Fig.16.42
Reinforcement details of Going and Landing
Now we will discuss some details about the above fig.
As in previous examples, bar ‘a’ becomes top bar in the top landing, and to compensate for this, bar ‘b’ is provided at the same diameter and spacing as ‘a’. Also, ‘a’ and ‘b’ are given enough embedment as explained earlier on the basis of fig.16.25. At the bottom intermediate landing, ‘c’ and ‘d’ are provided to resist any possible hogging moment. When there is hogging moment, these bars will be in tension, and so in order to prevent them from straightening up, they are given enough embedment in accordance with fig.16.25.
One more type of bar, named as ‘e’ is provided at the top landing. The reason for providing these bars can be explained as follows: The stair is supported on the landings. The support for the flight comes from the interior portion of the Landing. But the inner edge of the Landing (the line of intersection between horizontal landing portion and inclined flight portion) is also capable of providing some support. So the edge will act as an intermediate support of a continuous system. We know that hogging moments will develop at intermediate supports, and so top steel will have to be provided. Thus ‘e’ is provided to resist this hogging moment.
Another point that we have to note is the position of the main bars of the Landings. We can see that it is provided as the bottom most layer. This will give maximum effective depth, and so the ‘Moment of resistance’ MuR of the section of the Landing will be more. But in any case , these bars should indeed be the bottom most layer because, they have to provide support to the stair by carrying the bars(‘a’, ‘b’, ‘c’, ‘d’ and ‘e’) of the stairs. And in order to carry them, the bars of the Landing should be the bottom most layer.
Distributor bars for the Landing need not be designed because the bars of the stairs are coming in a transverse direction to the main bars of the Landing.
A small amount of steel (#8 @ 200 c/c) is also provided as top bars for the Landings. These bars will resist any hogging moment developed at the supports (the hatched portions in fig.16.41 above) of the Landings, and will also assist in tying the top bars of stairs, and thus to keep them in position.
It may also be noted that in the above fig., bars of different types are shown separately only for clarity. This was explained earlier based on the fig.16.25(b).
In the next section, we will discuss the load distribution in a particular type of longitudinal stairs.
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